Question

Heat is supplied to an ideal monoatomic gas $A$ so that it expands without changing its temperature. In another process, starting with the same state, heat is supplied at constant pressure. In both the cases, a graph of work done by the gas
$\left(W\right)$ is plotted versus heat added $\left(Q\right)$ to the gas. The ratio of the slopes of the graphs obtained in the first and second processes is ${\eta }_1$. The same ratio obtained for an ideal diatomic gas is ${\eta }_2$ Find the ratio $\frac{{\eta }_1}{{\eta }_2}$.

• A. $\frac{5}{2}$
• B. $\frac{5}{7}$
• C. $\frac{7}{2}$
• D. $\frac{3}{7}$

Answer 1

The correct option is B $\frac{5}{7}$

First law of thermodynamics says,
$Q=W+\Delta U$
For an isothermal process, $\Delta U=0$
$\therefore Q=W$
Thus, for the first process, we can say that the graph of $W$ vs $Q$ is a straight line with a positive slope $m_1=1$


For second (isobaric) process:
$\frac{Q}{\Delta U}=\frac{n{C}_p\Delta T}{n{C}_v\Delta T}=\gamma$
$\Rightarrow \frac{Q}{Q-W}=\gamma$
$\Rightarrow W=\frac{\gamma -1}{\gamma }Q=\frac{2}{5}Q$
$[\because \gamma =\frac{5}{3}\text{for monoatomic gas}]$
Therefore, slope of $W$ vs $Q$ graph for second (isobaric) process is $m_2=\frac{2}{5}$
$\therefore {\eta }_1=\frac{m_1}{m_2}=\frac{5}{2}$

Similarly, for a diatomic gas $(\gamma =\frac{7}{5})$
$m_1=1$ and $m_2=\frac{2}{7}$
$\therefore {\eta }_2=\frac{m_1}{m_2}=\frac{7}{2}$
Thus, $\frac{{\eta }_1}{{\eta }_2}=\frac{5/2}{7/2}=\frac{5}{7}$
Thus, option (b) is the correct answer.